U= "or" which means to add ∩ = and means to find common ,A U B' = A or not B because there's an or we add A and anything not B, if it was A ∩ B' , "A and not B" we find the common shaded regions in the two and only mark that
Because, A ∪ B' = { x | x ∈ A or x ∈ B' } = (A - B) ∪ (A ∩ B) ∪ B' Proof: LHS = A ∪ B' = ( A ∩ U ) ∪ B' ----- (• using Law of U) = ( A ∩ (B' ∪ B) ) ∪ B' ----- (• using Complement law) = [ (A ∩ B') ∪ (A ∩ B) ] ∪ B' = [ (A - B) ∪ (A ∩ B) ] ∪ B' ------ [• (A ∩ B') = (A - B) ] = ( A - B) ∪ (A ∩ B) ∪ B' = RHS
Because, A ∪ B' = (A - B) ∪ (A ∩ B) ∪ B' ⇒(A - B), (A ∩ B) and B' must be shaded to show A ∪ B' Proof: Solving LHS = A ∪ B' = [ A ∩ U ] ∪ B' = [ A ∩ (B' ∪ B) ] ∪ B' = [ (A ∩ B') ∪ (A ∩ B) ] ∪ B' = [ (A - B) ∪ (A ∩ B) ] ∪ B' = (A - B) ∪ (A ∩ B) ∪ B' = RHS
intersection is the common which is ( upside down u ) , and the union which is ( u ) is both sets together , and the apostrophe is like x bar which is a compliment , when the bar is on B it means u have to insert everything but B
